\( -\x \)

\( \vec{R}_1 = (-a,y,0) \), \( \vec{R}_2 = (a,y,0) \)

The vector \( \vec{R} \) is defined as \( \vec{R}_i = \vec{r} - \vec{r}_i \). Thus, \( \vec{R}_1 = (0,y,0) - (a,0,0) = (-a,y,0) \). (Notice that \( y \) is the reference point, that is, where we calculate the field.) Similarly, \( \vec{R}_2 = (0,y,0) - (-a,0,0) ) = (a,y,0) \).

\( \vec{R}_1 = (0,y,z) \), \( \vec{R}_2 = (2a,y,z) \)

In this case we find \( \vec{R}_1 = (a,y,z) - (a,0,0) = (0,y,z) \), and \( \vec{R}_2 = (a,y,z) - (-a,0,0) = (2a,y,z) \).

\( \vec{E} = Q/(4 \pi \epsilon_0) \left( (0,y,z)/(y^2 + z^2)^{3/2} - (2a,y,z)/(4 a^2 + y^2 + z^2 )^{3/2} \right) \)

The electric field is $$ \begin{equation} \vec{E} = \frac{1}{4 \pi \epsilon_0} \left[ Q \frac{(0,y,z)}{(y^2 + z^2)^{3/2}} - Q \frac{(2a,y,z)}{( 4 a^2 + y^2 + z^2 )^{3/2} } \right] \; . \tag{1.1} \end{equation} $$

Cylindrical around \( x \).

You can see from the system or the solution above that cylindrical symmetry around the \( x \)-axis would simplify the expressions. In this case, \( \rho = (y^2 + z^2) \) is the distance from the \( x \)-axis, \( \phi \) is the angle around the \( x \)-axis and \( x \) is the position along the cylinder axis.

\( \vec{E} = Q/(4 \pi \epsilon_0) \left[ \left( (x-a) \x + \rho \rhohat \right)/((x-a)^2 + \rho^2)^{3/2} - \left( (x+a)\x + \rho \rhohat \right) ( (x+a)^2 + \rho^2 )^{3/2} \right] \)

For the general case, \( \vec{R}_1 = (x-a,y,z) \) and \( \vec{R}_2 = (x+a,y,z) \). The electric field is: $$ \begin{equation} \vec{E} = \frac{1}{4 \pi \epsilon_0} \left[ Q \frac{(x-a,y,z)}{((x-a)^2 + y^2 + z^2)^{3/2}} - Q \frac{(x+a,y,z)}{( (x+a)^2 + y^2 + z^2 )^{3/2} } \right] \; , \tag{1.2} \end{equation} $$ which we can simplify by introducing \( \vec{\rho} = \rho \rhohat = (0,y,z) \), thus giving $$ \begin{equation} \vec{E} = \frac{1}{4 \pi \epsilon_0} \left[ Q \frac{(x-a) \x + \rho \rhohat }{((x-a)^2 + \rho^2)^{3/2}} - Q \frac{ (x+a)\x + \rho \rhohat }{( (x+a)^2 + \rho^2 )^{3/2} } \right] \; , \tag{1.3} \end{equation} $$

\( \rho_l = Q/L \)

We have to assume that the charge is uniformly distributed and that the rod is thin (has a small radius) so that it can approximated as a line. In this case, \( \rho_l = Q/L \)

\( \d q = (Q/L) \d x \)

\( \d q = \rho_l \d x = (Q/L) \d x \)

\( \d\vec{E} = \rho_l \d x'/(4 \pi \epsilon_0) ( -x',y,0)/ \left( (x')^2 + y^2 \right)^{3/2} \), \( \vec{R} = (-x',y,0) \).

In this case we use \( (x,y,z) \) for the position of the reference point. It is therefore useful to denote the position of the element \( \d x' \) as \( x' \), so that the position of the \( \d x' \) element is \( (x',0,0) \). We do this to avoid confusion with the reference position \( \vec{r} = (x,y,z) \). The contribution to the electric field is $$ \begin{equation} \d \vec{E} = \frac{\rho_l \d x'}{4 \pi \epsilon_0} \frac{\vec{R}}{R^3} \; . \tag{1.4} \end{equation} $$ In this case \( \vec{R} = \vec{r} - (x',0,0) = (0,y,0) - (x',0,0) = (-x',y,0) \).

\( \vec{E} = \int_{-L/2}^{L/2} \rho_l/(4 \pi \epsilon_0) (-x',y,0)/((x')^2 + y^2)^{3/2} \, \d x' \)

The electric field is found from the integral over all the charge contributions \( dq = \rho_l dx \): $$ \begin{equation} \vec{E} = \int_{-L/2}^{L/2} \frac{\rho_l \d x'}{4 \pi \epsilon_0} \frac{(-x',y,0)}{((x')^2 + y^2)^{3/2}} \; . \tag{1.5} \end{equation} $$ Notice that the integral is over \( x' \) and not \( x \), that is, \( x \) is a constant in the integration!

In this case, the symmetry of the object is cylinder symmetry along the \( x \)-axis. In addition, when we calculate the field along the \( y \)-axis, we notice that there will be pairs of contributions from \( -x' \) and \( x' \), and that the contributions from these pairs will cancel in the \( x \)-direction. Thus the field will only have a \( y \)-component at \( (0,y,0) \). The cylindrical symmetry means that we can use the result from the point \( (0,y,0) \) for any point in the \( yz \)-plane with the same distance to the \( x \)-axis.

\( E_x = 8.987 \text{kN/C} \)

In this case \( \vec{r} = (0,0) \) and \( \vec{R} = \vec{r} - \vec{r}_1 = (0,0) - (1,0) = (-1,0) = - 1 \x \) in units of meters. The electric field in the origin is is therefore $$ \begin{equation} \vec{E}(0,0) = \frac{1 \mu\text{C}}{4 \pi \epsilon_0} \frac{-(1 \text{m}) \x}{(1 \text{m})^3}\; , \tag{1.6} \end{equation} $$ The numerical value is

import numpy as np
import matplotlib.pyplot as plt
epsilon0 = 8.854187817e-12
q1 = 1e-6
R = -1
Ex = q1/(4.0*np.pi*epsilon0)*R/abs(R)**3
print(Ex)

-8987.55178799791

That is, the field is \( E_x = -8.987 \text{kN/C} \).

The function is

import numpy as np
import matplotlib.pyplot as plt
def efieldq(q1,r,r1):
    # Input: charge q1 in Coulomb
    #        r: position to find field (in 1,2 or 3 dimensions) in meters
    #        r1: position of charge q1 in meters
    # Output: electric field E at position r in N/C
    Rvec = r-r1
    Rnorm = np.sqrt(Rvec.dot(Rvec))
    epsilon0 = 8.854187817e-12
    E = q1/(4.0*np.pi*epsilon0)*Rvec/Rnorm**3
    return E

We test it by

q1 = 1e-6
r1 = np.array([1,0])
r = np.array([0,0])
E = efieldq(q1,r,r1)
print(E)

[-8987.551788     0.      ]

which is the same as we found in the exercise above.

\( \vec{r} = (a \, i, a \, j) \)

\( \vec{r} = (a \, i, a \, j) \)

\( \vec{r}_{i,j} = ((i-N)a,(j-N)a) \)

We write \( \vec{r}_{i,j} = (x_{i,j},y_{i,j}) \). We find \( x_{i,j} = -Na + ia = (i-N)a \) and \( y_{i,j} = -Na + ja = (j-N)a \)

\( \vec{F}_2 = (3.85\cdot 10^{-8},-1.92\cdot 10^{-8})\text{N} \)

We can solve this analytically by first realizing that \( \vec{R} \) is the vector from \( q_1 \) to \( q_2 \), \( \vec{R} = (0,1)-(2,1)\text{m} = (-2,1)\text{m} \). The force on \( q_2 \) is therefore: $$ \begin{eqnarray} \vec{F}_2 &=& \frac{q_1 q_2}{4 \pi \epsilon_0} \frac{\vec{R}}{R^3} \\ &=& \frac{4 \cdot 10^{-9} \cdot (-6) \cdot 10^{-9} \text{C} }{4 \pi 8.85 \cdot 10^{-12} \text{C}^2 \text{N}^{-1} \text{m}^{-2} } \frac{(-2,1)}{\sqrt{5}^3} \\ &=& (3.85\cdot 10^{-8},-1.92\cdot 10^{-8})\text{N} \; . \end{eqnarray} $$

Or we can solve this computationally:

import numpy as np
epsilon0 = 8.854187817e-12 # C^2/(N m^2)
q1 = 4e-9 # C
q2 = -6e-9 # C
r1 = np.array([2,0])
r2 = np.array([0,1])
R = r2-r1
F2 = q1*q2/(4*pi*epsilon0)*R/np.linalg.norm(R)**3
print("F2 = ",F2)

F2 =  [3.85858114e-08 -1.92929057e-08]

\( \vec{F}_2 = (-3.85\cdot 10^{-8},1.92\cdot 10^{-8})\text{N} \)

We know from Newton's third law that \( \vec{F}_1 = - \vec{F}_2 \).

We use the programs introduced in the textbook:

import numpy as np
import matplotlib.pyplot as plt
#
import numpy as np
import matplotlib.pyplot as plt
#
def efieldlist(r,Q,R):
    # Find E*4*pi*epsilon0 at r from a charge q at position r0
    E = np.zeros(np.shape(r))
    for i in range(len(R)):
        Ri = r - R[i]
        qi = Q[i]
        Rinorm = np.linalg.norm(Ri)
        E = E + qi*Ri/Rinorm**3
    return E
#
def findfield(R,Q,L,N):
    x = np.linspace(-L,L,N)
    y = np.linspace(-L,L,N)
    rx,ry = np.meshgrid(x,y)
    # Set up electric field
    Ex = np.zeros((N,N),float)
    Ey = np.zeros((N,N),float)
    # Calculate the field
    for i in range(len(rx.flat)):
        r = np.array([rx.flat[i],ry.flat[i]])
        Ex.flat[i],Ey.flat[i] = efieldlist(r,Q,R)
    return rx,ry,Ex,Ey
#
def visfield(rx,ry,Ex,Ey):
    Emag = np.sqrt(Ex**2 + Ey**2)
    minlogEmag = min(np.log10(Emag.flat))
    scaleE = np.log10(Emag) - minlogEmag
    uEx = Ex / Emag
    uEy = Ey / Emag
    # Visualize using both arrows and colors
    plt.figure(figsize=(10,5))
    ax1 = plt.subplot(1,2,1)
    plt.quiver(rx,ry,uEx*scaleE,uEy*scaleE)
    ax1.set_aspect('equal', 'box')
    ax2 = plt.subplot(1,2,2)
    plt.streamplot(rx,ry,Ex,Ey)
    ax2.set_aspect('equal', 'box')
    return

# Define charges
Q3 = []
R3 = []
R3.append(np.array([0,0]))
Q3.append(1.0)
# Visualize
rx,ry,Ex,Ey = findfield(R3,Q3,5,21)
visfield(rx,ry,Ex,Ey)

We calculate \( |E(x)| \) for \( x>0 \) and plot \( \log(|E|) \) as a function of \( x \). The slope in this log-log plot will be \( -2 \) if \( E(r) \propto r^{-2} \). This is done with the following script:

Q3 = []
R3 = []
R3.append(np.array([0,0]))
Q3.append(1.0)
x = np.linspace(0,100,1000)
E = np.zeros(len(x))
for i in range(len(x)):
    r = np.array([x[i],0.0])
    Ex,Ey = efieldlist(r,Q3,R3)
    E[i] = np.sqrt(Ex*Ex + Ey*Ey)
    j = np.where(np.log10(x)>0.0)
lx = np.log10(x[j])
ly = np.log10(abs(E[j]))
plt.plot(lx,ly,'o')
pol = np.polyfit(lx, ly, 1)
lfit = np.poly1d(pol)
llx = np.log10(x[i])
plt.plot(llx,lfit(llx),'-r')
print(pol)

[-2.00000000e+00 -6.77475917e-16]

The resulting behavior is shown in the figure.

\( E(r) \propto r^{-3} \)

We use the same approach as above, fitting in the range j = where(log10(x)>0.35). The resulting fit gives

[-3.02768353  0.65045551]

which means that the effective power-law is \( E(r) \propto r^{-3} \) for this dipole system. As we expected from theory.

\( E(r) \propto r^{-2} \)

If both charges have the same sign, the system will far away look like a system with one charge \( 2q \) in the origin, which will give a behavior \( E(r) \propto r^{-2} \).

\( E(r) \propto r^{-4} \)

\( Q/(2 \pi^2 \epsilon_0 a^2) \y \)

The charge of a small piece of length \( dl = a d \phi \) is \( dq = (Q/L) a d \phi = Q/(\pi a) a d \phi = Q/\pi d\phi \). The position is \( \vec{r}' = (a \cos \phi, a \sin \phi, 0) \) and \( \vec{r} = (0,0,0) \) so that \( \vec{R} = \vec{r} - \vec{r}' = (-a \cos \phi, - a \sin \phi,0) \) and \( R = a \). The contribution to the electric field from this piece is: $$ \begin{equation} d \vec{E} = \frac{dq}{4 \pi \epsilon_0} \frac{(-a \cos \phi, -a \sin \phi, 0)}{a^3} \tag{1.7} \end{equation} $$ We find the electric field by integrating from \( \phi = 0 \) to \( \pi \), getting $$ \begin{equation} \vec{E} = \frac{Q/\pi}{4 \pi \epsilon_0 a^2} \int_0^{\pi} (-\cos \phi, -\sin \phi, 0) d \phi = \frac{Q/\pi}{4 \pi \epsilon_0 a^2} \int_0^{\pi} (-\cos \phi, -\sin \phi, 0) d \phi = -\frac{2Q}{4 \pi^2 \epsilon_0 a^2} \y \tag{1.8} \end{equation} $$

The contribution to the charge from a small piece \( d \phi \) at \( \phi \) is \( dq = \rho_l(\phi) a d \phi = Q/(2a) a \sin \phi d \phi = Q/2 \sin \phi d \phi \). The integral from \( \phi = 0 \) to \( \phi = \pi \) is: $$ \begin{equation} \int_0^{\pi} \frac{Q}{2} \sin \phi d \phi = \frac{Q}{2}2 = Q \; . \tag{1.10} \end{equation} $$

\( \vec{E} = -\frac{Q}{16 \epsilon_0 a^2} \)

We find the magnitude and direction by integrating but not including the new expression for \( \rho_l(\phi) \). The integral becomes: $$ \begin{eqnarray} \vec{E} &=& \frac{1}{4 \pi \epsilon_0 a^2} \int_0^{\pi} (Q/2) \sin \phi (-\cos \phi, -\sin \phi, 0) d \phi \\ &=& \frac{Q}{8 \pi \epsilon_0 a^2} \int_0^{\pi} (-\sin \phi \cos \phi, -\sin^2 \phi, 0) d \phi \end{eqnarray} $$ We solve this integral using Sympy:

import sympy as sy
t = sy.Symbol('t')
sy.integrate(sy.sin(t)**2,(t,0,sy.pi))

The result is \( \pi/2 \). The field is therefore: $$ \begin{equation} \vec{E} = -\frac{Q}{8 \pi \epsilon_0 a^2} \frac{\pi}{2}\y = -\frac{Q}{16 \epsilon_0 a^2} \; . \tag{1.11} \end{equation} $$ We see that the electric field is largest for the varying charge density in part b .

\( \d \vec{E} = \rho_A r \d r \d \theta / (4 \pi \epsilon_0) (-r \cos \theta, - r\sin \theta, z)/(r^2 + z^2)^{3/2} \)

The distance from the element to the observation point \( (0,0,z) \) is \( R = \sqrt{r^2 + z^2} \), which is independent of \( \theta \). The position of the element is \( \vec{r}' = (r \cos \theta, r \sin \theta, 0) \). The \( \vec{R} \)-vector is therefore \( \vec{R} = \vec{r} - \vec{r}' = (0,0,z) - (r\cos \theta, r\sin \theta, 0) = (-r \cos \theta, - r\sin \theta, z) \). The area of the element is \( \d A = r \d r \d \theta \). The contribution from this element to the electric field is therefore: $$ \begin{equation} \d \vec{E} = \frac{\rho_A \d A}{4 \pi \epsilon_0} \frac{\vec{R}}{R^3} = \frac{\rho_A r \d r \d \theta}{4 \pi \epsilon_0} \frac{(-r \cos \theta, - r\sin \theta, z)}{(r^2 + z^2)^{3/2}} \; . \tag{1.12} \end{equation} $$

\( \vec{E} = \rho_A/(2 \epsilon_0) \left( 1 - z/(a^2 + z^2)^{1/2} \right) \z \)

We integrate the contributions from all the elements from \( r = 0 \) to \( r=a \) and from \( \theta = 0 \) to \( \theta = 2 \pi \). We notice that the integral of \( \sin \theta \) and \( \cos \theta \) is 0. Therefore, there will only be a contribution in the \( z \)-direction. $$ \begin{eqnarray} \vec{E} &=& \int_0^{a} \int_0^{2 \pi }\frac{\rho_A r \d r \d \theta}{4 \pi \epsilon_0} \frac{(-r \cos \theta, - r\sin \theta, z)}{(r^2 + z^2)^{3/2}} \d r \d \theta \\ &=& \frac{2 \pi \rho_A z}{4 \pi \epsilon_0} \int_0^{a} \frac{r}{(r^2 + z^2)^{3/2}} \d r \z \; . \end{eqnarray} $$

import sympy as sy
r = sy.Symbol('r')
a = sy.Symbol('a')
f = r/(r**2 + z**2)**(3/2)
sy.integrate(f,(r,0,a))

The result is \( 1/z - 1/(z^2 + a^2)^{1/2} \). The electric field is therefore: $$ \begin{equation} \vec{E} = \frac{\rho_A z}{2 \epsilon_0} \left( \frac{1}{z} - \frac{1}{(a^2 + z^2)^{1/2}} \right) \z\; . \tag{1.13} \end{equation} $$

\( \vec{E} = \rho_A/(2 \epsilon_0) \z \)

When \( z \rightarrow 0^{+} \) we see that \( \vec{E} = \rho_A/(2 \epsilon_0) \z \). However, the similar limit when \( z \rightarrow 0^{-} \) points in the opposite direction.

We introduce a dimensionless notation by introducing \( u = z/a \), rewriting the field to be $$ \begin{equation} E_z = \frac{\rho_A}{2 \epsilon_0} \left( 1 - \frac{z}{(z^2 + a^2)^{3/2}} \right) = \frac{\rho_A}{2 \epsilon_0} \left( 1 - \frac{(z/a)}{((z/a)^2 + 1)^{3/2}} \right) \; . \tag{1.14} \end{equation} $$ and plot the results:

u = np.linspace(0,10,1000)
E = (1-u/(u**2 + 1)**1.5
plot(u,E)

The resulting plot is shown in the figure.

Vi lager først en tegning av systemet, som viser linjen langs \( x \)-aksen. Vi tegner også inn punktet \( \vec{r} = (x,y) \) i figuren. Vi ser av figuren at for denne linjen svarer \( r \) til \( y \) og \( \rhat \) til \( \y \) når \( y>0 \) og \( -\y \) når \( y < 0 \). Det elektriske feltet er derfor $$ \begin{equation} \vec{E} = \frac{-(Q/L)}{2 \pi \epsilon_0 y} \y \; . \tag{1.16} \end{equation} $$ Merk at denne måte å skrive det på også gir riktig retning for feltet både når \( y>0 \) og når \( y < 0 \).

Først tegner vi en ny figur som viser begge linjene. Det eletriske feltet fra den nye linjen vil ha samme form som i forrige oppgave, men vi må nå ta eksplisitt hensyn til retningen til feltet.

Det elektriske feltet fra den nye linjen vil avhenge av avstanden til linjen gjennom \( y=a \). Det vil si at \( r \) i dette tilfellet vil være \( |y-a| \). Men så må vi passe på retningen til feltet. I dette tilfellet vil retningsvektor også peke i \( y \)-retningen, men den vil peke i positiv \( y \)-retning når \( y>a \) dvs. når \( y-a>0 \) og i negativ \( y \)-retning når \( y < a \) dvs når \( y-a < 0 \). Vi kan derfor skrive det elektriske feltet fra denne linjen som $$ \begin{equation} \vec{E} = \left\{ \begin{array}{cc} \frac{Q}{2 \pi \epsilon_0 L} \frac{1}{|y-a|}\y & \text{ når } y>a \\ \frac{Q}{2 \pi \epsilon_0 L} \frac{1}{|y-a|}(-\y) & \text{ når } y < a \end{array} \right. \tag{1.17} \end{equation} $$ Dette kan vi skrive mer kompakt ved å bruke at \( (y-a)/|y-a| \) er positiv når \( y>a \) og negativ når \( y < a \): $$ \begin{equation} \vec{E} = \frac{Q}{2 \pi \epsilon_0 L} \frac{(y-a)}{|y-a|^2}\y \; . \tag{1.18} \end{equation} $$ Det totale feltet er summen av de to feltene.

Fordi feltet ikke varierer langs \( x \)-aksen er det tilstrekkelig å plotte feltet langs \( y \)-aksen. Det gjør vi ved:

import numpy as np
import matplotlib.pyplot as plt
def E1(y):
    return 1/y
def E2(y,a):
    return (y-a)/(y-a)**2
a = 1.0
y = np.linspace(-a,2*a,1000)
E = E1(y)+E2(y,a)
plt.plot(y,E,'-k')
plt.ylim((-20,20))
plt.xlabel('$x/a$')
plt.ylabel('$E/E_0$')

Vi finner igjen feltet med superposisjonsprinsippet. For å finne et uttrykk for feltet for linjen langs \( y \)-aksen erstatter vi \( y \) med \( x \) i uttrykket vi fant over slik at $$ \begin{equation} \vec{E}_1 = \frac{Q}{2 \pi \epsilon_0 L} \frac{1}{y}\y \; . \tag{1.19} \end{equation} $$ og $$ \begin{equation} \vec{E}_2 = \frac{-Q}{2 \pi \epsilon_0 L} \frac{1}{x}\x \; . \tag{1.20} \end{equation} $$ Og det samlede feltet blir $$ \begin{equation} \vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{Q}{2 \pi \epsilon_0 L} (-1/x,1/y) \; . \tag{1.21} \end{equation} $$ For å plotte dette i \( xy \)-planet må vi lage en figur som viser både retningen og størrelsen på feltet. Vi bruker et vektor-plot for å vise dette:

import numpy as np
import matplotlib.pyplot as plt 
# Felt fra linje langs x-aksen
def E1(r,q):
    return q*np.array([0,1/r[1]])
# Felt fra linje langs y-aksen
def E2(r,q):
    return q*np.array([1/r[0],0])
Lx = 5
Ly = 5
N = 21
x = np.linspace(-Lx,Lx,N)
y = np.linspace(-Ly,Ly,N)
rx,ry = np.meshgrid(x,y)
# Set up electric field
Ex = np.zeros((N,N),float)
Ey = np.zeros((N,N),float)
# Calculate the field
for i in range(len(rx.flat)):
    r = np.array([rx.flat[i],ry.flat[i]])
    Ex.flat[i],Ey.flat[i] = E1(r,1.0)+E2(r,-1.0)
# Calculate field magnitude and unit vectors
Emag = np.sqrt(Ex**2 + Ey**2)
minlogEmag = np.nanmin(np.log10(Emag.flat))
scaleE = np.log10(Emag) - minlogEmag
uEx = Ex / Emag
uEy = Ey / Emag
# Visualize using both arrows and colors 
plt.figure(figsize=(8,4))
ax1 = plt.subplot(1,2,1)
scaleE = 1.0
plt.quiver(rx,ry,Ex,Ey)#uEx*scaleE,uEy*scaleE)
ax1.set_aspect('equal', 'box')
ax2 = plt.subplot(1,2,2)
plt.quiver(rx,ry,uEx,uEy,np.log10(Emag)) 
ax2.set_aspect('equal', 'box')

\( \rho_s(\vec{r}) = \rho_{s,0} \)

It means that the charge is the same everywhere in space.

Infinite

It will be infinite, since the plane is infinitely large and has a constant charge surface density.

We start from a given reference point \( (x,y,z) \). The horizontal contribution from a small surface element \( \d x' \d y' \) to the field at \( (x,y,z) \) will be cancelled by another element placed symmetrically around the point \( (x,y,0) \) as illustrated in the figure. The field will therefore only have a \( z \)-component.

\( \vec{R} = (x-x',y-y',z) \), \( \d \vec{E} = \rho_s \d x' \d y'/(4 \pi \epsilon_0) \, \vec{R}/R^3 \)

We use \( x' \) and \( y' \) for the element because we want to integrate over all elements. To avoid confusion with the coordinates \( (x,y,z) \) of the reference point, we use \( x' \) and \( y' \) for the surface element.

In this case, we find \( \vec{R} = (x,y,z) - (x',y',0) = (x-x',y-y',z) \). The contribution to the electric field from an element \( \d x' \d y' \) is: $$ \begin{equation} \d \vec{E} = \frac{\rho_s \d x' \d y'}{4 \pi \epsilon_0}\frac{\vec{R}}{R^3} \; . \tag{1.22} \end{equation} $$

\( \vec{E} = \int_0^a \int_0^b \rho_A/(4 \pi \epsilon_0) (x-x',y-y',z)/\left( (x-x')^2 + (y-y')^2 + z^2 \right)^{3/2} \d y' \d x' \)

We find the integral by inserting \( \vec{R} \) in the expression we found above: $$ \begin{equation} \vec{E} = \int_0^a \int_0^b \frac{\rho_A}{4 \pi \epsilon_0} \frac{(x-x',y-y',z)}{\left( (x-x')^2 + (y-y')^2 + z^2 \right)^{3/2} } \d y' \d x'\; . \tag{1.23} \end{equation} $$

0

Since the charges are placed symmetrically, all have the same magnitude, and all point in toward the center, the net force is zero.

\( \vec{F} = Qq/(4 \pi \epsilon_0 L^2) \hat{u} \), where \( \hat{u} \) points from the center to the removed charge.

We know that the sum of the five remaining charges and the removed charge is zero: \( \vec{F}_5 + \vec{F} = 0 \), which means that \( \vec{F}_5 = - \vec{F} \). The net force is therefore of the same magnitude as from a single charge, but in the opposite direction, pointing outwards if \( Qq>0 \). The magnitude of the force is \( F = Qq/(4 \pi \epsilon_0 L^2) \).

0 and \( Q/(4 \pi \epsilon_0 r^2) \x \)

The reasoning is there identical to the previous exercise. The net force is zero in the center of the circle when all the charges are placed symmetrical around the circle. Let us assume that a charge \( Q \) at \( \vec{r} = r\x \) is removed. The electric field is then the opposite of the electric field from the charge that is removed $$ \begin{equation} \vec{E} = \frac{Q}{4 \pi\epsilon_0 r^2} \x ; . \tag{1.24} \end{equation} $$

A spherical shell with outer radius \( a \) and inner radius \( b \) plus a sphere of radius \( b \) is a sphere of radius \( a \).

\( \vec{E}_a = \vec{E}_{\text{shell}} + \vec{E}_b \).

\( \vec{E}_{\text{shell}} = 0 \) for \( r < b \); \( \vec{E}_{\text{shell}} = \rho/(3 \epsilon_0) \left( r - b^3/r^2 \right) \rhat \) for \( b < r < a \).

The electric field from a uniformly charged sphere of radius \( a \) is the sum of the electric fields of a uniformly charged sphere of radius \( b \) and a uniformly charged spherical shell, all with the same volume charge densities: \( \vec{E}_a = \vec{E}_{\text{shell}} + \vec{E}_b \) as illustrated in the figure. This gives that \( \vec{E}_{\text{shell}} = \vec{E}_a - \vec{E}_b \).

Inside the empty center of a spherical shell, that is, for \( r < b \), we know that \( \vec{E}_a = \rho/(3 \epsilon_0) r \rhat \) and \( \vec{E}_b = \rho/(3 \epsilon_0) r\rhat \) so that \( \vec{E}_{\text{shell}} = 0 \).

Inside the shell, that is, for \( b < r < a \), we know that \( \vec{E}_a = \rho/(3 \epsilon_0) r \rhat \) and \( \vec{E}_b = \rho/(3 \epsilon_0) (b^3/r^2)\rhat \) so that: $$ \begin{equation*} \vec{E}_{\text{shell}} = \vec{E}_a - \vec{E}_b = \frac{\rho}{3 \epsilon_0} r \rhat - \frac{\rho}{3 \epsilon_0} \frac{a^3}{r^2} \rhat = \frac{\rho}{3 \epsilon_0} \left( r - \frac{b^3}{r^2} \right) \rhat \; . \end{equation*} $$

\( W = Q^2/(2 \pi \epsilon_0 a) \)

The system is illustrated in the figure. When charge A is in a position \( y \) the \( \vec{R} \)-vector from charge B is \( \vec{R}_B = (-a/2,y) \) and from C is \( \vec{R}_C = (a/2,y) \). The net force from the two charges on charge A is therefore $$ \begin{equation} \vec{F} = \frac{Q^2}{4 \pi \epsilon_0} \frac{(-a/2,y)}{R^3} + \frac{Q^2}{4 \pi \epsilon_0} \frac{(a/2,y)}{R^3} = \frac{Q^2}{4 \pi \epsilon_0} \frac{2y}{R^3} \hat{y} \; . \tag{1.25} \end{equation} $$ The work done to carry out the displacement must be a force in the opposite direction to \( \vec{F} \). The displacement is from \( h = \sqrt{3}/2 \) to \( y = 0 \). The work is therefore: $$ \begin{equation} W = \int_h^0 - \vec{F} \cdot dy \hat{y} = \int_h^0 -\frac{Q^2}{4 \pi \epsilon_0} \frac{2y}{R^3} dy = -\frac{Q^2}{4 \pi \epsilon_0} \int_h^0 \frac{2y}{(y^2 + (a/2)^2)^{3/2}}dy \tag{1.26} \end{equation} $$ We solve this using Sympy:

import sympy as sy
a = sy.Symbol('a')
y = sy.Symbol('y')
F = 2*y/(y**2 + (a/2)**2)**(3/2)
sy.integrate(F,(y,sqrt(3)/2*a,0))

2/a

The work is therefore $$ \begin{equation} W = \frac{Q^2}{4 \pi \epsilon_0}\frac{2}{a}=\frac{Q^2}{2 \pi \epsilon_0 a} \; . \tag{1.27} \end{equation} $$

Superposition.

\( E_z = \rho_s/(2 \epsilon_0) \, z/(z^2 + a^2)^{1/2} \)

The plan is to solve this by first finding the field from the infinite plane and then subtracting the field from a disk of radius \( a \). We can do this by calculating the field from a disk with radius \( a \). The infinite field is found by letting \( a \rightarrow \infty \). To find the field from a disk or radius \( a \), we integrate the contributions from elemental rings of radius \( r \) and width \( dr \). The field from such a ring is directed in the \( z \)-direction due to symmetry, so we only look at the \( z \)-components, \( E_z = \vec{E} \cdot \hat{z} \): $$ \begin{equation} d \vec{E} = \frac{dQ}{4 \pi \epsilon_0} \frac{\vec{R}}{R^3} \; , \tag{1.28} \end{equation} $$ where \( dQ \) is the charge of the disk element, \( dQ = \rho_s dA \), where \( dA = 2 \pi r dr \) is the area of the disk element. We find the \( z \)-component: $$ \begin{equation} dE_z = \frac{2 \pi r dr \rho_s}{4 \pi \epsilon_0} \frac{\vec{R}\cdot \hat{z}}{R^3} = \frac{r dr\rho_s}{2 \epsilon_0} \frac{z }{R^3} \; , \tag{1.29} \end{equation} $$ where \( R = (z^2 + r^2)^{1/2} \). The electric field from a finite plate of size \( a \) is therefore: $$ \begin{equation} E_z = \int_0^a \frac{r dr\rho_s}{2 \epsilon_0} \frac{r z dr}{(z^2 + r^2)^{3/2} } = \frac{z \rho_s}{2 \epsilon_0} \int_0^a \frac{r}{(z^2 + r^2)^{3/2} } \, dr \tag{1.30} \end{equation} $$ We solve this using Sympy:

import sympy as sy
r = sy.Symbol('r')
z = sy.Symbol('z')
a = sy.Symbol('a')
f = r/(r**2 + z**2)**(3/2)
sy.integrate(f,(r,0,a))

The result is \( 1/z - 1/(z^2 + a^2)^{1/2} \). The electric field is therefore: $$ \begin{equation} E_z = \frac{z \rho_s}{2 \epsilon_0} \left( \frac{1}{z} - \frac{1}{(z^2 + a^2)^{1/2}} \right) = \frac{\rho_s}{2 \epsilon_0} \left( 1 - \frac{z}{(z^2 + a^2)^{1/2}} \right) \; . \tag{1.31} \end{equation} $$ The field from an infinite plate is the limit when \( a \rightarrow \infty \), which is $$ \begin{equation} E_{z,\infty} = \frac{\rho_s}{2 \epsilon_0} \; . \tag{1.32} \end{equation} $$ The field from the plate with the hole is the field from the plate minus the field of the hole: $$ \begin{eqnarray} E_z &=& E_{z,\infty} - E_{z,a} \\ &=& \frac{\rho_s}{2 \epsilon_0} - \frac{\rho_s}{2 \epsilon_0} \left( 1 - \frac{z}{(z^2 + a^2)^{1/2}} \right) \\ &=& \frac{\rho_s}{2 \epsilon_0} \frac{z}{(z^2 + a^2)^{1/2}} \; . \end{eqnarray} $$

\( \sum \vec{F} = 0 \)

The torque is \( \vec{\tau} = \vec{r}\times\vec{F} \), where \( \vec{r} \) is the the vector going from the center of mass to where the force \( \vec{F} \) is applied.

Using the formula for torque and cross product rules we get $$ \begin{eqnarray} \vec{\tau} &=& \vec{r} \times \vec{F} = \left(\frac{\vec{d}}{2} \times q\vec{E}\right) + \left(-\frac{\vec{d}}{2} \times (-q\vec{E})\right) \\ &=& q\vec{d} \times \vec{E} = \vec{p}\times\vec{E} \end{eqnarray} $$

Stable equilibrium: \( \theta = 0 \), Unstable equilibrium: \( \theta = \pi \), maximum torque: \( \theta = \frac{\pi}{2} \ \ \wedge \ \ \theta = \frac{3\pi}{2} \)

A simple harmonic oscillator has the differential equation $$ \begin{equation} m\frac{\d^2 x}{\d t^2} = -kx \; , \tag{1.33} \end{equation} $$ where \( m \) and \( k \) are constants and \( x \) describes the harmonic motion. The solution is $$ \begin{equation} x(t) = Acos(\omega t + \phi) \quad , \quad \omega = \sqrt{\frac{k}{m}} \; , \tag{1.34} \end{equation} $$ where \( A \) and \( \phi \) are constants.

Looking at the torque: $$ \begin{equation} \vec{\tau} = \vec{p}\times\vec{E} \implies \tau = -|\vec{p}||\vec{E}|sin(\theta) = I\alpha \tag{1.35} \end{equation} $$ Using the small angle approximation we get $$ \begin{equation} I\alpha \approx -|\vec{p}||\vec{E}|\theta \tag{1.36} \end{equation} $$ This gives us the differential equation $$ \begin{equation} I\frac{\d^2\theta}{\d t^2} = -|\vec{p}||\vec{E}|\theta \tag{1.37} \end{equation} $$ This is an equation for a simple harmonic oscillator.

We are looking for a solution like $$ \begin{equation} \theta(t) = Acos(\omega t + \phi) \tag{1.39} \end{equation} $$ In order for energy to be conserved \( A = \theta_0 \) where \( \theta_0 \) is the maximum value of \( \theta \). \( \omega \) is known and \( \phi \) depends on the initial conditions, lets say the \( \theta \) peaks at \( t=0 \), then \( \phi = 0 \). The angle \( \theta \) can therefore be described by the equation $$ \begin{equation} \theta(t) = \theta_0 cos\left(\sqrt{\frac{|\vec{p}||\vec{E}|}{I}}t\right) \tag{1.40} \end{equation} $$

The formula for moment of inertia for point mass $$ \begin{equation} I = \sum_i m_ir_i² \tag{1.41} \end{equation} $$ \( m_i \) is the mass and \( r_i \) the distance from the point of rotation to the point mass.

There is an electrostatic force pointing upwords and gravitational force down. They are the same size.

Negative

\( q = 4.8 \times 10^{-19}\text{C} \)

3

\( \vec{E}_z = \frac{\rho}{4\epsilon_0}\z \)

The contribution to the electric field from a small surface element is given by $$ \begin{equation} \d\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{\rho}{R^2}\rhat \; , \tag{1.42} \end{equation} $$ where \( \rhat \) is a unit vector pointing from the surface element to the origin. Notice that because the origin is equidistant to the hemisphere, the distance from any surface element to the origin is the same. Because of symmetry we only need to look at the component in \( z \)-direction (the \( x \)- and \( y \)-components cancel out). Therefore we get the following expression: $$ \begin{equation} \d E_z = \d\vec{E}\cdot\z = \frac{1}{4\pi\epsilon_0}\frac{\rho}{R^2}\rhat \cdot \z = \frac{1}{4\pi\epsilon_0}\frac{\rho}{R^2}cos(\theta) \tag{1.43} \end{equation} $$ Make sure you understand the last transition, if not; express \( \rhat \) in spherical coordinates and then carry out the dot product. Because we are working with a spherical shape it can be a good idea to transition to spherical coordinates when we do the surface integral. Substituting \( \d S = R^2sin(\theta)\d\phi \d\theta \) we get the following integral: $$ \begin{equation} E_z = \frac{\rho}{4\pi \epsilon_0} \int_0^{2\pi}\int_0^{\pi/2}sin(\theta)cos(\theta)\d\theta \d\phi = \frac{\rho}{4\epsilon_0} \tag{1.44} \end{equation} $$ Therefore we get the field $$ \begin{equation} \vec{E}_z = \frac{\rho}{4\epsilon_0}\z \; . \tag{1.45} \end{equation} $$